At takeoff a commercial jet has a 45.0 m/s speed. Its tires have a diameter of 0.850 m.
(a) At how many rpm are the tires rotating?
Incorrect: Your answer is incorrect.
rpm
(b) What is the centripetal acceleration at the edge of the tire?
4764.70
Correct: Your answer is correct.
m/s2
(c) With what force must a determined 10-15 kg bacterium cling to the rim?
N
(d) Take the ratio of this force to the bacterium's weight.
(force from part (c) / bacterium's weight)
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HEYYYY!!!!
MOMMMMMMAAAA!!!
a. Circumference = pi*D = 3.14 * 0.85 = 2.67 m.
1rev/2.67m * 45m/s * 60s/min = 1012 rev/min.
IT'S LITTTT!!!!!!
To solve part (a) of the problem, we need to convert the speed of the jet in meters per second (m/s) to the rotational speed of the tires in revolutions per minute (rpm).
First, let's calculate the number of revolutions the tires make per second. We can find this by dividing the speed of the jet by the circumference of the tires.
The formula for the circumference of a circle is given by:
Circumference = 2πr
Given that the diameter of the tires is 0.850 m, the radius (r) can be calculated by dividing the diameter by 2:
r = 0.850 m / 2 = 0.425 m
Now, let's calculate the circumference:
Circumference = 2π(0.425 m) = 2.67 m
The number of revolutions per second can be calculated by dividing the speed of the jet by the circumference:
Revolutions per second = 45.0 m/s / 2.67 m/rev = 16.85 rev/s
Now, let's convert the revolutions per second to revolutions per minute:
Revolutions per minute (rpm) = 16.85 rev/s * 60 s/min = 1011 rpm
Therefore, the tires are rotating at approximately 1011 rpm.
For part (b) of the problem, we are asked to find the centripetal acceleration at the edge of the tire. The centripetal acceleration (a) can be calculated using the formula:
a = v^2 / r
where v is the linear velocity and r is the radius of the tire.
Given that the speed of the jet is 45.0 m/s and the radius of the tire is 0.425 m (half the diameter), we can substitute these values into the formula:
a = (45.0 m/s)^2 / 0.425 m = 4764.70 m/s^2 (rounded to the nearest hundredth)
Therefore, the centripetal acceleration at the edge of the tire is approximately 4764.70 m/s^2.
For part (c) of the problem, we need to find the force required for a bacterium to cling to the rim of the tire. This force can be found using the formula:
Force = mass x centripetal acceleration
Given that the mass of the bacterium is 10^-15 kg and the centripetal acceleration we calculated in part (b) is 4764.70 m/s^2, we can substitute these values into the formula:
Force = (10^-15 kg) x (4764.70 m/s^2) = 4.76 x 10^-12 N
Therefore, the force required for the bacterium to cling to the rim of the tire is approximately 4.76 x 10^-12 Newtons.
For part (d) of the problem, we need to calculate the ratio of this force to the bacterium's weight. The weight of the bacterium can be found using the formula:
Weight = mass x gravitational acceleration
Given that the mass of the bacterium is 10^-15 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute these values into the formula:
Weight = (10^-15 kg) x (9.8 m/s^2) = 9.8 x 10^-15 N
Now, let's calculate the ratio:
Ratio = (force from part (c)) / (bacterium's weight) = (4.76 x 10^-12 N) / (9.8 x 10^-15 N) = 4.86 x 10^2
Therefore, the ratio of the force required for the bacterium to cling to the rim of the tire to the bacterium's weight is approximately 4.86 x 10^2.